Question

Write an equation of a line in slope-intercept form that is perpendicular to the line 2x -3y = 12 and passes through the point (2, 6).

Answer 1

For this case we have by definition, that the equation of a line in the slope-intersection form is given by:

`y = mx + b`

Where:

m: It's the slope

b: It is the cutoff point with the y axis

We have the following line:

`2x-3y = 12\\2x-12 = 3y\\y = \frac {2} {3} x-4`

If the line we wish to find is perpendicular to the one given, then its slope is given by:

`m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {\frac {2} {3}}\\m_ {2} = - \frac {3} {2}`

Then the line is:

`y = - \frac {3} {2} x + b`

We substitute the point:

`6 = - \frac {3} {2} (2) + b\\6 = -3 + b\\b = 6 + 3\\b = 9`

Finally, the equation is:

`y = - \frac {3} {2} x + 9`

Answer:

`y = - \frac {3} {2} x + 9`